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<H2>Time Dilation: A Worked Example </H2>

<P>
<I>Michael Fowler</I> </P>
<P>
<I>UVa Physics</I><BR>
</P>
<A HREF="lecturelist.html">Index of Lectures and Overview of the Course</A><BR>
<A HREF="synchronizing.html">Link to Previous Lecture</A>
<BR>

<H3>&quot;Moving Clocks Run Slow&quot; <I>plus</I> &quot;Moving
Clocks Lose Synchronization&quot; <I>plus</I> &quot;Length Contraction&quot;
leads to consistency! </H3>

<P>
The object of this exercise is to show explicitly how it is possible
for two observers in inertial frames moving relative to each other
at a relativistic speed to each see the other's clocks as running
slow and as being unsynchronized, and yet if they both look at
the same clock at the same time from the same place (which may
be far from the clock), they will <I>agree</I> on what time it
shows!</P>
<P>
Suppose that in frame <I>S</I> we have two synchronized clocks
<I>C<SUB>1</SUB></I> and <I>C<SUB>2</SUB></I> set 18 x 10<SUP>8</SUP>
meters apart (that's about a million miles, or 6 light-seconds).
A spaceship carrying a clock <I>C'</I> is traveling at 0.6<I>c</I>,
that is 1.8 x 10<SUP>8</SUP> meters per second, parallel to the
line <I>C<SUB>1</SUB>C<SUB>2</SUB></I>, passing close by each
clock.</P> 
<P>
<IMG SRC="img00020.gif" HEIGHT=234 WIDTH=716><BR>
</P>
<P>
Suppose <I>C'</I> is synchronized with <I>C<SUB>1</SUB></I> as
they pass, so both read zero.<BR>
</P>
<P>
As measured by an observer <I>O</I> in <I>S</I> - the &quot;ground
frame&quot; --<I> </I>the spaceship will take just 10 seconds
to reach <I>C<SUB>2</SUB></I>, since the distance is 6 light seconds,
and the ship is traveling at 0.6<I>c</I>.<BR>
</P>
<P>
What does clock <I>C</I>' (the clock on the ship) read as it passes
<I>C<SUB>2</SUB></I>? <BR>
</P>
<P>
The time dilation factor <IMG SRC="img00022.gif" HEIGHT=32 WIDTH=151>,
so <I>C</I>', the ship's clock, will read 8 seconds.<BR>
</P>
<P>
Thus if both <I>O</I>, <I>O'</I> are at <I>C<SUB>2</SUB></I> as
<I>C'</I> passes <I>C<SUB>2</SUB></I> , both will agree that the
clocks look like:<BR>
<IMG SRC="img00021.gif" HEIGHT=133 WIDTH=104><BR>
</P>
<P>
<I><B>How, then, can O' claim that the clocks C<SUB>1</SUB>, C<SUB>2</SUB>
are the ones that are running slow?<BR>
</P></B></I>
<P>
To <I>O'</I>, <I>C<SUB>1</SUB></I>, <I>C<SUB>2</SUB></I> <I>are</I>
running slow, but remember they are <I>not synchronized</I>. To
<I>O'</I>, <I>C<SUB>1</SUB></I> is <I>behind</I> <I>C<SUB>2</SUB></I>
by <IMG SRC="img00023.gif" HEIGHT=26 WIDTH=283> seconds. 
<P>
Therefore, <I>O'</I> will conclude that since <I>C<SUB>2</SUB></I>
reads 10 seconds as she passes it, at that instant <I>C<SUB>1</SUB></I>
must be registering 6.4 seconds. <I>O' </I>'s own clock reads
8 seconds at that instant, <I>so she concludes that C<SUB>1</SUB>
is running slow by the appropriate time dilation factor of 4/5</I>.
This is how the change in synchronization makes it possible for
both <I>O</I> and <I>O'</I> to see the other's clocks as running
slow. 
<P>
Of course, <I>O</I>'s assertion that as she passes the second
&quot;ground&quot; clock <I>C</I><SUB>2</SUB> the first &quot;ground&quot;
clock <I>C</I><SUB>1</SUB> must be registering 6.4 seconds is
not completely trivial to check! After all, that clock is now
a million miles away!<BR>
</P>
<P>
Let us imagine, though, that both observers are equipped with
Hubble-style telescopes attached to fast acting cameras, so reading
a clock a million miles away is no trick. 
<P>
To settle the argument, the two of them agree that as she passes
the second clock, the ground observer will be stationed at the
second clock, and at the instant of her passing they will both
take telephoto digital snapshots of the faraway clock <I>C<SUB>1</SUB></I>,
to see what time it reads. 
<P>
<I>O</I>, of course, knows that <I>C<SUB>1</SUB></I> is 6 light
seconds away, and is synchronized with <I>C<SUB>2</SUB></I> which
at that instant is reading 10 seconds, so his snapshot must show
<I>C<SUB>1</SUB></I> to read 4 seconds. That is, looking at <I>C</I><SUB>1</SUB>
he sees it as it was six seconds ago.<BR>
</P>
<P>
<A NAME="OLE_LINK1">What does </A><I>O' </I>'s digital snapshot
show? It must be identical -- two snapshots taken from the same
place at the same time must show the same thing! So, <I>O'  must
also</I> gets a picture of <I>C<SUB>1</SUB></I> reading 4 seconds.

<P>
<I><B>How can she reconcile a picture of the clock reading 4 seconds
with her assertion that at the instant she took the photograph
the clock was registering 6.4 seconds? <BR>
</P></B></I>
<P>
The answer is that she can if she knows her relativity!
<P>
<I>First point: length contraction</I>. To <I>O'</I>, the clock
<I>C<SUB>1</SUB></I> is actually only 4/5 x 18 x 10<SUP>8</SUP>
meters away (she sees the distance <I>C<SUB>1</SUB>C<SUB>2</SUB></I>
to be Lorentz contracted!).
<P>
 <I>Second point: The light didn't even have to go that far!</I>
In her frame, the clock <I>C<SUB>1</SUB></I> is <I>moving away</I>,
so the light arriving when she's at <I>C<SUB>2</SUB></I> must
have left <I>C<SUB>1</SUB></I> when it was closer -- at distance
<I>x </I>in the figure below.<I> </I>The figure shows the light
in her frame moving from the clock towards her at speed <I>c</I>,
while at the same time the clock itself is moving to the left
at 0.6<I>c</I>. </P>

<P>It might be helpful to imagine yourself in her frame of reference, 
so you are at rest, and to think of clocks <I>C<SUB>1</SUB></I> and 
<I>C<SUB>2</SUB></I> as being at the front end and back end respectively of a train 
that is going past you at speed 0.6<I>c</I>. Then, at the moment 
the back of the train passes you, you take a picture (through your telescope, of course) 
of the clock at the front of the train. Obviously, the light from the front clock 
that enters your camera at that instant left the front clock some time ago. 
During the time that light traveled towards you at speed <I>c</I>, the front of the 
train itself was going in the opposite direction at speed 0.6<I>c</I>.
But you know the length of the train in your frame is 4/5 x 18 x 10<SUP>8</SUP> meters, 
so since at the instant you take the picture the back of the train is passing you, 
 the front of the train must be 4/5 x 18 x 10<SUP>8</SUP> meters away. Now that 
distance, 4/5 x 18 x 10<SUP>8</SUP>, is the sum of the distance the light entering 
your camera traveled plus the distance the train traveled in the same time, that is, (1 + 0.6)/1 times the distance the light traveled. </P>

<P>So the image of the first ground clock  she sees and records
as she passes the second ground clock must have been emitted when
the first clock was a distance<I> x</I> from her in her frame,
where <I>x</I>(1 + 3/5) = 4/5 x 18 x 10<SUP>8</SUP> m., so <I>x</I>
= 9 x 10<SUP>8</SUP> meters.</P>
<P>
Having established that the clock image she is seeing as she takes
the photograph left the clock when it was only 9 x 10<SUP>8</SUP>
meters away, that is, 3 light seconds, she concludes that she
is observing the first ground clock as it was three seconds ago.
</P>
<P>
<IMG SRC="img00024.gif" HEIGHT=207 WIDTH=693><BR>
</P>
<P>
<I>Third point: time dilation</I>.  The story so far: she has
a photograph of the first ground clock that shows it to be reading
4 seconds. She knows that the light took three seconds to reach
her.  So, what can she conclude the clock must actually be registering
at the instant the photo was taken?  If you are tempted to say
7 seconds, you have forgotten that in her frame, the clock is
moving at 0.6<I>c</I> and hence <I>runs slow</I> by a factor 4/5.

<P>
Including the time dilation factor correctly, she concludes that
in the 3 seconds that the light from the clock took to reach her,
the clock itself will have ticked away 3 x 4/5 seconds, or 2.4
seconds. 
<P>
Therefore, since the photograph shows the clock to read 4 seconds,
and she finds the clock must have run a further 2.4 seconds, she
deduces that at the instant she took the photograph the clock
must actually have been registering 6.4 seconds, which is what
she had claimed all along!</P>
<P>
The key point of this lecture is that at first it seems impossible
for two observers moving relative to each other to both maintain
that the other one's clocks run slow. However, by bringing in
the other necessary consequences of the theory of relativity,
the Lorentz contraction of lengths, and that clocks synchronized
in one frame are out of synchronization in another by a precise
amount that follows necessarily from the constancy of the speed
of light, the whole picture becomes completely consistent!</P>
</P><A HREF="sreltwins.html"><I>Link to Next Lecture</I></A>
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<P>

Copyright &copy; Michael Fowler, 1996</P> 
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